Michael's triangle

Geometry Level 4

In quadrilateral ABCD,ABCD, AB=AD=1,AB = AD = 1, BD=2,BD = \sqrt{2}, BC=22,BC = \dfrac{ \sqrt{2} }{ 2 }, and CD=62.CD = \dfrac{ \sqrt{6} }{ 2}. The area of triangle ABCABC can be expressed in the form a+bc \dfrac{\sqrt{a} + b}{c} , where a,a, b,b, and cc are positive integers and aa is square-free. What is the value of a+b+ca+b+c?

This problem is posed by Michael T.

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